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Simplifying Trigonometric Identities>
Simplifying Trigonometric IdentitiesSimplifying Trigonometric Identities
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Question 1 of 7
1. Question
Which of the following are trigonometric identities?- 1.
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2.
cot2θ=cos2θ+1cot2θ=cos2θ+1 -
3.
sec2θ=1+sin2θsec2θ=1+sin2θ -
4.
sec2θ=1+tan2θsec2θ=1+tan2θ
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Trigonometric Identities
secθ=1cosθsecθ=1cosθcscθ=1sinθcscθ=1sinθcotθ=1tanθcotθ=1tanθtanθ=sinθcosθtanθ=sinθcosθcotθ=cosθsinθcotθ=cosθsinθUsing the given image, we can see that the radius of the circle will be 11. Therefore, the circle follows the formula x2+y2=1x2+y2=1, where 11 is the radius.Make a reference triangle by making a line from the point of origin (0,0)(0,0) going to any point in the circle to represent the radius and connect it to either the x or y-axisNotice that, given angle θθ, the circle has the vertical line parallel to the y-axis as its opposite side and the line on the x-axis as the adjacent sideNow that we have the opposite and adjacent sides and the radius 11 as the hypotenuse, we can use the trigonometric functions to find their respective valuessinsin == oppositehypotenuseoppositehypotenuse == y1y1 == yy coscos == adjacenthypotenuseadjacenthypotenuse == x1x1 == xx xx and yy are both coordinate values, so we can write the coordinates as (cosθ,sinθ)(cosθ,sinθ)Finally, use the values of xx and yy and derive its trigonometric identitiesx2+y2x2+y2 == 11 cos2θ+sin2θcos2θ+sin2θ == 11 Substitute known values (cos2θ+sin2θ)(cos2θ+sin2θ)÷cos2θ÷cos2θ == 11÷cos2θ÷cos2θ Divide both sides by cos2θcos2θ 1+sin2θcos2θ1+sin2θcos2θ == 1cos2θ1cos2θ 1+1+tan2θtan2θ == 1cos2θ1cos2θ sin2θcos2θ=tan2θsin2θcos2θ=tan2θ 1+tan2θ1+tan2θ == sec2θsec2θ 1cos2θ=sec2θ1cos2θ=sec2θ sec2θsec2θ == 1+tan2θ1+tan2θ x2+y2x2+y2 == 11 cos2θ+sin2θcos2θ+sin2θ == 11 Substitute known values (cos2θ+sin2θ)(cos2θ+sin2θ)÷sin2θ÷sin2θ == 11÷sin2θ÷sin2θ Divide both sides by sin2θsin2θ cos2θsin2θ+1cos2θsin2θ+1 == 1sin2θ1sin2θ cot2θcot2θ+1+1 == 1sin2θ1sin2θ cos2θsin2θ=cot2θcos2θsin2θ=cot2θ cot2θ+1cot2θ+1 == csc2θcsc2θ 1sin2θ=csc2θ1sin2θ=csc2θ csc2θ = cot2θ+1 sec2θ=1+tan2θcsc2θ=cot2θ+1 -
Question 2 of 7
2. Question
Simplify11+sinθ+11-sinθ-
1.
csc2θ+2 -
2.
csc2θ -
3.
2sec2θ -
4.
2sec2θ
Hint
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Trigonometric Identities
secθ=1cosθcscθ=1sinθcotθ=1tanθtanθ=sinθcosθcotθ=cosθsinθApply the rule of adding fractions and simplify the expression11+sinθ+11-sinθ = 1-sinθ+1+sinθ(1+sinθ)(1-sinθ) = 2(1+sinθ)(1-sinθ) Combine like terms = 21-sin2θ Expand Recall that sin2θ+cos2θ=1. Derive this formula to further simplify the expressionsin2θ+cos2θ = 1 sin2θ+cos2θ -sin2θ = 1 -sin2θ Subtract sin2θ from both sides cos2θ = 1-sin2θ 21-sin2θ = 2cos2θ 1-sin2θ=cos2θ = 21×1cos2θ Take out 2 = 2×sec2θ 1cos2θ=sec2θ = 2sec2θ 2sec2θ -
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Question 3 of 7
3. Question
Simplifytanθsecθ-
1.
1tan θ -
2.
sin θ -
3.
2sec θ -
4.
cos θ
Hint
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Trigonometric Identities
secθ=1cosθcscθ=1sinθcotθ=1tanθtanθ=sinθcosθcotθ=cosθsinθSubstitute known values to tan and sectan = sincos sec = 1cos tan θsec θ = sincos1cos Substitute known values = sin θ Apply rule of dividing fractions sin θ -
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Question 4 of 7
4. Question
Simplifysecθ-secθsin2θ-
1.
csc θ -
2.
sin θ -
3.
sec θ -
4.
cos θ
Hint
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Trigonometric Identities
secθ=1cosθcscθ=1sinθcotθ=1tanθtanθ=sinθcosθcotθ=cosθsinθFactor out sec θ on both termssec θ-sec θ sin2θ = sec θ(1-sin2θ) Factorise Recall that sin2θ+cos2θ=1. Derive this formula to further simplify the expressionsin2θ+cos2θ = 1 sin2θ+cos2θ -sin2θ = 1 -sin2θ Subtract sin2θ from both sides cos2θ = 1-sin2θ sec θ(1-sin2θ) = sec θ(cos2θ) 1-sin2θ=cos2θ = 1cos θ⋅(cos2θ) sec θ=1cos θ = cos2θcos θ = cos θ cos θ -
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Question 5 of 7
5. Question
Simplifysec2A-csc2A-
1.
(sec A-csc A)(sec A-csc A) -
2.
(sec A-csc A)(sec A+csc A) -
3.
(sec A+csc A)(sec A+csc A) -
4.
A(sec A+csc A)
Hint
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The difference of two squares is a format of a quadratic expression where there are two terms that are both perfect squares.First, find the square of the two terms.First termsec2A = sec A⋅sec A Second termcsc2A = csc A⋅csc A Finally, get the sum and difference of the two values.(sec A-csc A)(sec A+csc A) (sec A-csc A)(sec A+csc A) -
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Question 6 of 7
6. Question
Simplifytan2θsecθ-1-
1.
secθ+1 -
2.
secθ-1 -
3.
-1secθ -
4.
1secθ
Hint
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Trigonometric Identities
secθ=1cosθcscθ=1sinθcotθ=1tanθtanθ=sinθcosθcotθ=cosθsinθRecall that 1+tan2θ=sec2θ. Derive this formula to further simplify the expression1+tan2θ = sec2θ 1+tan2θ -1 = sec2θ -1 Subtract 1 from both sides tan2θ = sec2θ-1 tan2θsecθ-1 = sec2θ-1secθ-1 tan2θ=sec2θ-1 = (secθ-1)(secθ+1)secθ-1 Factorise = secθ+1 secθ-1secθ-1=1 secθ+1 -
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Question 7 of 7
7. Question
Simplify(tan2β)(cot2β+1)tan2β+1- (1)
Hint
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Trigonometric Identities
secθ=1cosθcscθ=1sinθcotθ=1tanθtanθ=sinθcosθcotθ=cosθsinθRecall that 1+cot2θ=csc2θ. Derive this formula to further simplify the expression(tan2β)(cot2β+1)tan2β+1 = tan2β⋅csc2θtan2β+1 1+cot2θ=csc2θ Finally, substitute known values to tan, csc, and sectan = sincos csc = 1sin sec = 1cos tan2β⋅csc2θtan2β+1 = sin2βcos2β⋅1csc2θ1cos2θ Substitute known values = sin2βcos2β⋅1csc2θ⋅cos2θ1 Apply rule of dividing fractions = 1 1